<![CDATA[MTH721 Assignment 1 Solution and Discussion]]>17f5033a-06ba-402d-8c14-96c6a1ac78a4-image.png MTH721 (Spring 2020) Assignment No. 1

                                                                                         Maximum Marks: 25      
                                                                                       Due Date: May 31, 2020

INSTRUCTIONS

Please read the following instructions before attempting the solution of this assignment:
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• Try to get the concepts, consolidate your concepts which you learn in these lectures with
these questions.
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Note: Attempt all the following questions.

Question: 1 Marks: 5
Determine whether the binary operation * defined by :R×R→R and given for all a,b∈R as : ab=〖(a+b)〗^2 is associative or not? Explain your answer.

Question: 2 Marks: 5
Show that C, the set of all non-zero complex numbers is a multiplicative group.
Question: 3 Marks: 5
Show that the following function f:Z_2→Z_2 is a ring homomorphism:
f(x)=x^2
Question: 4 Marks: 5
Show that the following function g:Z→Z is not a ring homomorphism:
f(x)=2x
Question: 5 Marks: 5
Show that in a principal ideal domain, every nonzero prime ideal is maximal.

]]>https://community.secnto.com//topic/1916/mth721-assignment-1-solution-and-discussionRSS for NodeMon, 08 Jun 2026 21:54:05 GMTMon, 15 Jun 2020 08:40:46 GMT60<![CDATA[Reply to MTH721 Assignment 1 Solution and Discussion on Mon, 15 Jun 2020 08:44:22 GMT]]>@cyberian said in MTH721 Assignment 1 Solution and Discussion:

Question: 5 Marks: 5
Show that in a principal ideal domain, every nonzero prime ideal is maximal.

4c9b9d5e-bb02-46ca-832d-f900c32bd998-image.png

]]>https://community.secnto.com//post/5381https://community.secnto.com//post/5381Mon, 15 Jun 2020 08:44:22 GMT<![CDATA[Reply to MTH721 Assignment 1 Solution and Discussion on Mon, 15 Jun 2020 08:43:15 GMT]]>@cyberian said in MTH721 Assignment 1 Solution and Discussion:

Question: 4 Marks: 5
Show that the following function g:Z→Z is not a ring homomorphism:
f(x)=2x

457718aa-5b8f-4d07-aa63-4d622808d51f-image.png

]]>https://community.secnto.com//post/5380https://community.secnto.com//post/5380Mon, 15 Jun 2020 08:43:15 GMT<![CDATA[Reply to MTH721 Assignment 1 Solution and Discussion on Mon, 15 Jun 2020 08:42:47 GMT]]>@cyberian said in MTH721 Assignment 1 Solution and Discussion:

Question: 3 Marks: 5
Show that the following function f:Z_2→Z_2 is a ring homomorphism:
f(x)=x^2

a2b11f8a-e058-40c9-8c30-2d24e3ab4fde-image.png

]]>https://community.secnto.com//post/5379https://community.secnto.com//post/5379Mon, 15 Jun 2020 08:42:47 GMT<![CDATA[Reply to MTH721 Assignment 1 Solution and Discussion on Mon, 15 Jun 2020 08:42:09 GMT]]>@cyberian said in MTH721 Assignment 1 Solution and Discussion:

Question: 2 Marks: 5
Show that C, the set of all non-zero complex numbers is a multiplicative group.

65e25f18-ae7a-4cd7-9f31-720bb9c5d95c-image.png Answer:
Let C={z:z=x+iy, x,y∈R}C={z:z=x+iy, x,y∈R}. Here R is the set of all real numbers and i=√(-1).
(G1) Closure Axiom: If a+ib∈C and c+id∈C, then by the definition of multiplication of complex numbers
(a+ib)(c+id)=(ac–bd)+i(ad+bc)∈C
Since ac–bd,ad+bc∈R, for a,b,c,d∈R. Therefore,C is closed under multiplication.
(G2) Associative Axiom:
(a+ib){(c+id)(e+if)}=(ace–adf–bcf–bde)+i(acf+ade+bce–bdf)
={(a+ib)(c+id)}(e+if) for a,b,c,d∈R .
(G3) Identity Axiom: e=1(=1+i0) is the identity in C.
(G4) Inverse Axiom: Let (a+ib)(≠0)∈C, then
(a+ib)^(-1)=1/(a+ib)=(a-ib)/(a^2+b^2 )
=a/(a^2+b^2 )-i b/(a^2+b^2 )=m+in∈∁
Hence C is a multiplicative group.

]]>https://community.secnto.com//post/5378https://community.secnto.com//post/5378Mon, 15 Jun 2020 08:42:09 GMT<![CDATA[Reply to MTH721 Assignment 1 Solution and Discussion on Mon, 15 Jun 2020 08:41:40 GMT]]>@cyberian said in MTH721 Assignment 1 Solution and Discussion:

Question: 1 Marks: 5
Determine whether the binary operation * defined by :R×R→R and given for all a,b∈R as : ab=〖(a+b)〗^2 is associative or not? Explain your answer.

13d8ee93-13d1-4ef8-b584-a7f4641d7a0f-image.png Answer:
Consider the elements 1,3,6∈R. Then we have that:

1∗(2∗3)=1∗(2+3)2=1∗25=(1+25)2=676
We also have that:

(1∗2)∗3=(1+2)2∗3=9∗3=(9+3)2=122=144
Clearly 676≠144 and so ∗ is nonassociative on R since a∗(b∗c)≠(a∗b)∗c for 1,3,6∈R.

]]>https://community.secnto.com//post/5377https://community.secnto.com//post/5377Mon, 15 Jun 2020 08:41:40 GMT